Answer to Question #130489 in General Chemistry for Abhisek

Question

If the electrone falls from n=5 and n=4 in the H-atom then emitted energy is

Answer 1

To calculate the emitted energy use the following formula: $E_{mn}=13.6\;\rm eV \left(\dfrac{1}{n^2}-\dfrac{1}{m^2}\right)$

where n and m are the electron levels (m>n).

If n=4 and m=5, the energy is $E=13.6\;\rm eV\left (\dfrac{1}{4^2}-\dfrac{1}{5^2}\right)=0.306\;\rm eV\;or\;4.90\times10^{-20}\;\rm J$

(1 eV=1.60*10^(-19) J)

If n=1 and m=4 E=12.8 eV (2.05*10^(-18) J)

If n=1 and m=5 E=13.1 eV (2.10*10^(-18) J)